#### Answer

Any of:
$4\sin x\cos x(1-2\sin^{2}x)$
$4\sin x\cos x(\cos^{2}x-\sin^{2}x)$
$4\sin x\cos x(2\cos^{2}x- 1)$

#### Work Step by Step

$\sin 4x=\sin[2(2x)] =$
... Double-Angle identity
$...\sin 2A=2\sin A\cos A$
$=2\sin 2x\cos 2x$
... Double-Angle
... $\sin 2A=2\sin A\cos A$
... $\cos 2A=1-2\sin^{2}A$
$=2(2\sin x\cos x)(1-2\sin^{2}x)$
$=4\sin x\cos x(1-2\sin^{2}x)$
(answers may differ if other double angle identities for cosine are chosen,
$\cos 2A=\cos^{2}A-\sin^{2}A$,
$\cos 2A=2\cos^{2}A- 1$)
Answer: any of
$4\sin x\cos x(1-2\sin^{2}x)$
$4\sin x\cos x(\cos^{2}x-\sin^{2}x)$
$4\sin x\cos x(2\cos^{2}x- 1$